Problem: Find the greatest value of $b$ such that $-b^2+7b-10 \ge 0$.
Answer: We factor the quadratic, getting $(b-5)(2-b) \ge 0$. The expression is equal to $0$ when $b=5 \text{ or } 2$.  When $b \le 2$ or $b \ge 5$, the quadratic is negative. When $2 \le b \le 5$, the quadratic is non-negative. Therefore, the greatest value of $b$ for which $(b-5)(2-b)\ge 0$ is $b=\boxed{5}$.